Proof: Divisibility of Integers by 4

[Math100]

Homework Statement

Establish the following divisibility criteria:
An integer is divisible by $4$ if and only if the number formed by its tens and units digits is divisible by $4$.
[Hint: $10^{k}\equiv 0\pmod {4}$ for $k\geq 2$.]

Relevant Equations

None.

Proof:

Let $N$ be an integer.
Then $N=a_{m}10^{m}+a_{m-1}10^{m-1}+\dotsb +a_{1}10+a_{0}$ for $0\leq a_{k}\leq 9$.
Note that $10^{k}\equiv 0\pmod {4}$ for $k\geq 2$.
Thus $4\mid N\Leftrightarrow N\equiv 0\pmod {4}\Leftrightarrow a_{1}10+a_{0}\equiv 0\pmod {4}$.
Therefore, an integer is divisible by $4$ if and only if the number formed by its tens and units digits is divisible by $4$.

 

[fresh_42]

Correct, but why do you say integers and then restrict everything to positive numbers? The same is true for negative numbers.

 

[Math100]

Correct, but why do you say integers and then restrict everything to positive numbers? The same is true for negative numbers.

I guess I was a bit thoughtless. So should I mention "Let N be a natural number.", instead?

 

[dextercioby]

Is this exercise necessarily in "modular arithmetics", or can we solve it like in 5th grade (that is when kids are about 11 yo)?

 

[fresh_42]

I guess I was a bit thoughtless. So should I mention "Let N be a natural number.", instead?

No. Integers are fine. Everything remains true if you put a minus sign in front of $N.$

E.g., you can say: Let $N$ be an integer. We assume w.l.o.g. $N\geq 0$ because the negative case can be treated the same way.

w.l.o.g. stands for: without loss of generality. It means that the assumption is allowed because it is no real restriction.

 

[SammyS]

Proof:

Let $N$ be an integer.
Then $N=a_{m}10^{m}+a_{m-1}10^{m-1}+\dotsb +a_{1}10+a_{0}$ for $0\leq a_{k}\leq 9$.

You can write $N$ as:

$\displaystyle N=\left(a_{m}10^{m-2}+a_{m-1}10^{m-3}+\dotsb + a_{3}10+a_{2}\right) \,100 + a_{1}10+a_{0}$